Question

The general solution of the differential equation $\frac{dy}{dx} = \cot x \cdot \cot y$ is ______.

• A. $\cos x = c \csc y$
• B. $\sin x = c \sec y$
• C. $\sin x = c \cos y$
• D. $\cos x = c \sin y$

Hint:

Using $\int \tan \theta \,d\theta = -\ln|\cos \theta|$ is generally much safer than using $\ln|\sec \theta|$ during separation of variables, as it avoids complex fraction inversions later when removing the logarithms!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Use the Variable Separable Method: move all $y$ terms to one side and $x$ terms to the other.

Step 2: Formula Application:
$\frac{1}{\cot y} dy = \cot x \, dx \implies \tan y \, dy = \cot x \, dx$.

Step 3: Explanation:
Integrate both sides: $\int \tan y \, dy = \int \cot x \, dx$ $\log |\sec y| = \log |\sin x| + \log c$ $\log |\sec y| = \log |c \sin x|$ $\sec y = c \sin x \implies \sin x = \frac{1}{c} \sec y$.

Step 4: Final Answer:
The general solution is $\sin x = c \sec y$ (or $\sin x \cos y = C$).